Inequality $\frac{1-3ab}{1-2ac}+\frac{1-3bc}{1-2ba}+\frac{1-3ca}{1-2cb}\geq 0$ 
Let $a\ne 0$, $b\ne 0$ and $c\ne 0$ such that $a^2+b^2+c^2=1$. Prove that: $$\dfrac{1-3ab}{1-2ac}+\dfrac{1-3bc}{1-2ba}+\dfrac{1-3ca}{1-2cb}\geq 0.$$

My attempt to the solution: We get that $ab +bc+ca$ lies between $-0.5$ and $1$. We can use this. But I don't know how? 
 A: I will use the notation $\sum_{cyc} f(x,y,z)$ to mean that the sum is over the so-called cyclic permutations of $(x,y,z)$ (or the even ones if you prefer),
so $\sum_{cyc} f(x,y,z):=f(x,y,z)+f(y,z,x)+f(z,x,y)$ (in place of $x,y,z$ there will also be other variables).
Lemma. If $x,y,z\ge 0$ then $\sum_{cyc} (x+y-2z)(z+x)(x+y)\ge 0$. 
Proof. We have $\sum_{cyc}(x+z-2y)(z+x)(x+y)$ 
$=\sum_{cyc}(x^2z+xyz+x^3+x^2y+xz^2+yz^2+x^2z+xyz-2xyz-2y^2z-2x^2y-2xy^2)$ 
$=\sum_{cyc} x^3-2\sum_{cyc} x^2y+\sum_{cyc} xy^2=\sum_{cyc} x(x-y)^2\ge 0$. $\blacksquare$
Your inequality (using the hypothesis $a^2+b^2+c^2=1$) can be written as $\sum_{cyc}\frac{(a-b)^2+c^2-ab}{(a-c)^2+b^2}\ge 0$. 
Let us call $A=(b-c)^2+a^2$ and $B$, $C$ similarly (by cyclically permuting the variables).
 The inequality becomes $\sum_{cyc} \frac{C}{B}\ge\sum_{cyc}\frac{ab}{B}$. 
Putting $S=a^2+b^2+c^2$ we clearly have $ab=\frac{S-C}{2}$ (and the cyclic ones), so the inequality is equivalent to
$3\sum_{cyc}\frac{C}{B}\ge \sum_{cyc}\frac{S}{B}$ or (which is the same) $3\sum_{cyc}\frac{B}{A}\ge \sum_{cyc}\frac{S}{A}$. 
We will prove something slightly stronger, namely that $3\sum_{cyc}\frac{B}{A}\ge\sum_{cyc}\frac{A+B+C}{A}$ (this implies your inequality since $A+B+C\ge S$). 
Now observe that $A-B=2c(a-b)\le (a-b)^2+c^2=C$ (and by symmetry $B-C\le A$, $C-A\le B$), so that $A,B,C$ satisfy the triangle inequalities
$A\le B+C$, $B\le C+A$, $C\le A+B$. 
We can thus perform the substitution $A=y+z,B=z+x,C=x+y$ with $x,y,z\ge 0$ (if you didn't know this fact, check it!). 
So we are left to prove that $\sum_{cyc}\frac{3(z+x)-2(x+y+z)}{y+z}\ge 0$, which is the Lemma (upon clearing the positive denominators).
